Let the co-ordinates of one vertex of $\mathrm{△}\mathrm{ABC}\text{ be }\mathrm{A}(0,2,\mathrm{\alpha })$ and the other two vertices lie on the line $\frac{\mathrm{x}+\mathrm{\alpha }}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}.\text{ For }\mathrm{\alpha }\in \mathrm{\mathbb{Z}}$, if the area of  $\mathrm{△}\mathrm{ABC}$ is 21 sq. units and the line segment BC has length  $2\sqrt{2}$ units, then ${\mathrm{\alpha }}^2$ is equal to ______.

area of $∆\mathrm{ABC}=21$
$\begin{array}{l}\Rightarrow \frac{1}{2}\times \mathrm{BC}\times \mathrm{h}=21\\ \Rightarrow \mathrm{h}=\sqrt{21},\end{array}$
Let $\mathrm{D}=(5\mathrm{k}-\mathrm{\alpha },2\mathrm{k}+1,3\mathrm{k}-4)$

Drs of AD$\left(5k-\alpha , 2k-1, 3k-4-\alpha \right)$
Drs of BC are (5, 2, 3)
$\begin{array}{l}\mathrm{AD}\perp \mathrm{BC}\\ \Rightarrow 5(5\mathrm{k}-\mathrm{\alpha })+2(2\mathrm{k}-1)+3(3\mathrm{k}-4-\mathrm{\alpha })=0\\ \Rightarrow 25\mathrm{k}-5\mathrm{\alpha }+4k-2+9\mathrm{k}-12-3\mathrm{\alpha }=0\\ \Rightarrow 38\mathrm{k}-8\mathrm{\alpha }-14=0\\ \Rightarrow 19\mathrm{k}-4\mathrm{\alpha }-7=0\end{array}$
$\begin{array}{l}\mathrm{AD}=\sqrt{21}\\ \Rightarrow (5\mathrm{k}-\mathrm{\alpha }{)}^2+(2\mathrm{k}-1{)}^2+(3\mathrm{k}-4-\mathrm{\alpha }{)}^2=21\\ \Rightarrow 3{\mathrm{\alpha }}^2+20\mathrm{\alpha }-87=0\\ \Rightarrow \mathrm{\alpha }=3\left(\text{ or }\right)\mathrm{\alpha }=\frac{-29}{3}\\ \therefore {\mathrm{\alpha }}^2=9\end{array}$